4
$\begingroup$

This a probably very easy question and I am not sure whether it has been asked before (although I searched for it). Moreover I really hope this is nothing which can be found in any standard commutative algebra text book.


Is there a thorough discussion of Taylor expansions of polynomials (or maybe rational functions) with coefficients in an arbitrary field?


Problem: In the classical Taylor expansion we have coefficients of the form $\frac{1}{n!}$, which obviously don't have to exists in any field, say, of finite characteristic.

Motivation: I came across it at a discussion of the Zariski tangent space to a scheme $X$ at a point $y$. Say $X=\mathbb A_k^n$ and $y\in X$ corresponds to the maximal ideal $\frak m$. Then the differential $D_y$ induces an isomorphism $\frak {m}/\frak {m}^2\rightarrow E^\vee$, where $\frak E^\vee$ is the dual of the vector space $k^n$, thus identifies the Zariski tangent space with the "classical" tangent space as we know it from the theory of manifolds.

The prove usually involves something like: $\frak {m}=(T_1-\lambda_1,..., T_n-\lambda_n)$, then use the Taylor expansion of any polynomial $P$ in $(\lambda_1,...,\lambda_n)$.

Remark: I guess this is easy to fix for polynomials, but I wonder whether everything works in general for say rational functions, or anything fancier.

$\endgroup$
1

1 Answer 1

4
$\begingroup$

I found the answer myself - thanks to the comment by KConrad. Apparently everything can be fixed with the hasse derivative http://math.fontein.de/2009/08/12/the-hasse-derivative/. Thanks for your help.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.